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Chapter 20 Section 2. self-ionization - the reaction in which two water molecules produce ions H 2 O + H 2 O H 3 0 + + OH - hydroxide ion (OH - ) – what a water molecule that loses a hydrogen ion becomes - PowerPoint PPT Presentation
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Chapter 20 Section 2
self-ionization- the reaction in which two water molecules produce ions
H2O + H2O → H30+ + OH-
hydroxide ion (OH-) – what a water molecule that loses a hydrogen ion becomes
hydronium ion (H30+) – what a water molecule that gains a hydrogen ion becomes
-in pure water at 25°C the concentration of [H+] and [OH-] are 1.00 x 10-7mol/L
* [H+] = [OH-] in pure waterneutral solution- [H+] and [OH-] are equal
-in any aqueous solution, as [H+] inc. [OH-] dec. and vice versa
-in aqueous solutions: [H+] x [OH-] = 1.00 x 10-14M2
Kw = ion product constant for water Kw = [H+] x [OH-] = 1.0 x 10-14 M2
[H+] = Kw / [OH-]
[OH-] = Kw / [H+]
-not all solutions are neutral → some release H+ when dissolved in water
ex- HCℓ → H+ + Cℓ-
-such solutions have a greater [H+] than [OH-] acidic solution- [H+] is greater than [OH-]
-so [H+] > 1.0 x 10-7 M-some solutions release [OH-]ex- NaOH → Na+ + OH-
basic solution- [H+] is less than [OH-]-so [H+] < 1.0 x 10-7 M
**looking at negative exponent (ex -7 < -2)
-[H+] can also be expressed using pH scale
pH = potential hydrogen
-pH scale ranges from 0-14-pH < 7 = acids 0= strongly acidic-pH > 7 = bases 14= strongly basic-pH = 7 = neutral solution
**based on scales of 10-page 584 Table 20.2
pH = -log [H+]pOH = -log [OH-]pH + pOH = 14
Examples: [H+]= 1.0 x 10-7
pH = -log 1.0 x 10-7
pH = 7.00 neutral solution
[H+]= 4.3 x 10-3
pH = -log 4.3 x 10-3 pH = 2.37 acidic solution
• To find [H+] or [OH-] from a pH or pOH use:
[H+] = antilog - pH
[OH-] = antilog –pOH
antilog = 10x button
Ex: pH = 3.40Find [H+][H+] = antilog -3.40
= 3.98 x 10-4M